Mathematics Study Guide for the TABE Test

Algebraic Concepts: Additional Algebra Topics

Algebra has been a big part of this study guide so far. In this section, we will expand upon the ideas you’ve already learned to explain some slightly more complex algebraic concepts..

Rewriting Expressions

Often in algebra, expressions need to be rewritten to help solve a problem. One technique that can be useful in this process is factoring, which is when we remove a common factor out of two or more terms or expressions. You learned how to do this with numbers in an earlier section. Now you’ll learn how to factor out variables and terms.

For instance, in \(7x^2 +4x\), both terms have an \(x\). We can thus factor the \(x\) out of each term:

\[7x^2+4x\] \[x(7x)+x(4)\]

Per the distributive property, this can be rewritten as:

\[x(7x+4)\]

Another technique is called the difference of two squares. As the name suggests, this involves the subtraction of one square from another. If you see two perfect squares with a minus sign between them, say \(x^2-a^2\), they can be factored as \((x+a)(x-a)\).

For instance, in \(x^2-9\), both \(x^2\) and \(9\) are perfect squares, with one being subtracted from the other. In this case, the factors are \((x+3)(x-3)\).

Here’s another example with two variables:

\[4x^2-y^2 = (2x)^2 - y^2 = (2x+y)(2x-y)\]

As you can see, we needed an intermediary step to make the first term easier to break down, but once we did that, it was simple.

Let’s look at one more example that’s a bit more complicated. Does \(x^4-y^4\) fit the pattern? Well, all you need to determine is if both terms are perfect squares. This can be determined by rewriting the expression:

\[(x^2)^2 - (y^2)^2\]

Yes, the original two terms were perfect squares. So, using the factoring pattern, we get:

\[(x^2 + y^2)(x^2 - y^2)\]

Another trick comes up here. The second factor, \((x^2-y^2)\), is also the difference of two squares, so we can factor it as \((x+y)(x-y)\). Now we have three factors:

\[(x^2+y^2)(x+y)(x-y)\]

This can happen when the exponents are greater than two.

Polynomials

There are specific combinations of constants, coefficients, and variables that are known as polynomials. Depending on how many terms there are in a polynomial, we give them special names.

Monomials have one term, such as:

\[3x\] \[x^2\]

Binomials have two terms, such as:

\[81x^2+ 14\] \[11-5x\]

Trinomials have three terms, such as:

\[y^2-3y+74\] \[x^2 +xz + z^2\]

You can have an infinite number of terms in a polynomial, though four or five are the most you’re likely to see.

Polynomials as a System

Similar to basic numbers, you can add, subtract, and multiply polynomials. Polynomials are what is known as a closed system, meaning that when you do those basic operations with them, your result will also be a polynomial.

Note: When doing division with polynomials, sometimes the result includes a variable with a negative exponent, which is, by definition, not a polynomial. So, division doesn’t maintain the closed system.

It’s important to do two things when adding polynomials. First, make sure both polynomials have terms in the same order, usually in order of decreasing exponents. Second, line them up so that like terms are in the same column.

Let’s practice by adding \(5x^2 -2x +1\) and \(x^2+2x-2\):

\[\begin{array}{rrr} 5x^2 & -2x & +1\\ +x^2 &+2x & -2 \\ \hline 6x^2 &+ 0 &-1\\ \end{array}\]

To subtract would require the same setup, but of course you would be looking for the difference.

You can also multiply two polynomials using the distributive property:

\[(x-5)(x^2 -3x+9)\]

First, take the \(x\) from the left polynomial and multiply it by every term in \((x^2 -3x+9)\):

\[x(x^2) +x(-3x) +x(9)\]

Now, do the same with the \(-5\):

\[(-5)(x^2) +(-5)(-3x)+(-5)(9)\]

Put all six terms together and simplify:

\[x^3 -3x^2 + 9x -5x^2+15x-45\] \[x^3 -8x^2 +24x -45\]

Notice that in both cases we ended up with another polynomial.

Finding the Solutions of an Equation

When solving an equation, the goal is to find the value (or values) of the variable that make the equation true. These values are called the solutions of the equation.

For example, consider this linear equation:

\[2x+5=17\]

To solve it, we first isolate the variable:

\[2x=17 - 5\] \[2x = 12\]

Next, we divide both sides by \(2\) to get the value of \(x\):

\[\frac{2x}{2} = \frac{12}{2}\] \[x = 6\]

The solution is \(x=6\) because replacing \(x\) with \(6\) makes the equation true:

\[2(6)+5=17\]

Finding the Solutions of Quadratic Equations

Some equations can have more than one solution. This is especially common with *quadratic equations, which contain a variable raised to the second power. Quadratic equations are used in various real-world situations, such as determining the path of a thrown ball, finding the area of shapes, and calculating costs and revenues for a business .

A quadratic equation is usually written in this form:

\[ax^2 + bx + c = 0\]

where \(a\), \(b\), and \(c\) are numbers and \(a \neq 0\).

One common method for solving quadratic equations is factoring. You learned about factoring when rewriting expressions. In this situation, the idea is the same but the process is a little more involved.

Consider this equation:

\[x^2-5x+6=0\]

To factor this equation, we must look for two numbers that:

• when multiplied, equal \(6\)
• when added, equal \(-5\)

Doing a little mental math, we can see that those numbers are \(-2\) and \(-3\). Now, we can factor the equation as:

\[(x-2)(x-3)=0\]

Next, we will apply the zero product property, which states that if the product of two expressions is zero, then at least one of the expressions must equal zero.

So, we’ll set each factor equal to zero:

\[x-2=0 \qquad \text{or} \qquad x-3=0\]

Finally, we’ll solve each equation:

\[x=2 \qquad \text{or} \qquad x=3\]

So, the equation has two solutions:

\[x=2\qquad \text{ and }\qquad x=3\]

You can check the solutions by substituting them back into the original equation.

Finding the Zeros

Finding the zero of a function just means to set it equal to zero and solve for \(x\). In other words, the zero of a function is the value of \(x\) that makes the function equal to zero. Take the function \(y = 6x-18\). To find its zero, let’s turn into into an equation:

\[6x-18=0\] \[6x=18\] \[x=3\]

We say that \(3\) is a solution of the equation \(6x-18=0\) and a zero of the function \(y = 6x-18\).

Rational Expressions and Functions

A rational expression contains ratios with polynomial expressions in the top and bottom (numerator and denominator). Here are a few examples of rational expressions:

\[\frac{z-1}{z+1}\] \[\frac{x^2-4}{x+2}\] \[\frac{t^3-4t}{t^2-4t+4}\]

In the same way that you can simplify standard fractions, you need to be able to simplify rational expressions by factoring and canceling. Let’s practice this process by simplifying the second expression above:

\[\frac{x^2-4}{x+2}\]

First, we’ll factor the numerator:

\[\frac{(x+2)(x-2)}{x+2}\]

From here, we can cancel out entire expressions that are identical:

\[\require{cancel} \frac{\cancel{{(x+2)}}(x-2)}{\cancel{(x+2)}}\] \[x-2\]

If you want it to be a rational function, write \(f(x)=\) before the expression:

\[f(x) = \frac{x^2-4}{x+2}\]

Rational and Radical Equations

A rational equation is an equation that contains a rational expression, which means a variable may appear in a denominator. When solving a rational equation, remember that a denominator can never equal \(0\).

Consider this equation:

\[\frac{x^2-4}{x-2}=0\]

To solve, the numerator can be factored:

\[\frac{(x-2)(x+2)}{x-2}=0\]

The numerator equals \(0\) when:

\[x=2\]

or

\[x=-2\]

However, \(x=2\) would make the denominator equal \(0\), so it cannot be a solution.

Therefore:

\[x=-2\]

A radical equation is an equation in which a variable appears inside a radical. One common way to solve a radical equation is to isolate the radical and then raise both sides to the appropriate power.

Consider this equation:

\[\sqrt{x+5}=x-1\]

To solve, start by squaring both sides:

\[x+5=(x-1)^2\]

Now, you can expand:

\[x+5=x^2-2x+1\]

Move all terms to one side:

\[x^2-3x-4=0\]

At this point, you will factor:

\[(x-4)(x+1)=0\]

This gives two possible solutions:

\[x=4\]

or

\[x=-1\]

When a radical equation is solved by squaring both sides, the process can produce an extraneous solution. An extraneous solution is a value that appears during the solving process but does not satisfy the original equation. You need to check both solutions.

Start by checking \(x=4\):

\[\sqrt{4+5}=4-1\] \[3=3\]

So, \(x=4\) is a solution.

Now, check \(x=-1\):

\[\sqrt{-1+5}=-1-1\] \[2\neq-2\]

So, \(x=-1\) is an extraneous solution.

Always check possible solutions in the original equation, especially after raising both sides of an equation to a power.

Equations in Two or More Variables

While we’ve shown how equations with a single variable can be used to represent a word problem, real-world situations often have more than one unknown. You’ve learned about how to work with pairs of equations (or functions) with more than one variable. That same logic can be applied to a word problem.

For instance, imagine that Old McDonald has cows and pigs on his farm. In total, he has \(17\) animals. There are five fewer pigs than cows. Can you figure out how many cows there are?

We have two unknowns, so let’s give them both variables. Let the number of cows be \(c\) and the number of pigs be \(p\). Together they add up to \(17\):

\[c+p=17\]

We’re also told the difference between their numbers is \(5\), so we can create a second equation:

\[c-p=5\]

So, now we have simultaneous equations and we need to figure out the value of the two variables. You’ve learned how to do this. Just add the two equations:

\[\begin{array}{rcrcr} c &+&p&=&17\\ c &-&p&=&5\\ \hline & &2c&=&22\\ \hline & &c& =&11\\ \end{array}\]

The number of cows is \(11\).

Note: We used the elimination method here, but we could have also used the substitution method.

Graphing Equations in Two Variables

Just as you can graph an equation with one variable, you can graph an equation with two variables. There’s nothing really new about it. Just make up a data table and plot the points you get. Let’s create a graph of this equation below:

\[y = \frac{x^2-4}{2}\]

First, we’ll pick some values for \(x\) and calculate corresponding \(y\) values:

\[\begin{array}{|c|c|c|c|c|c|c|c|} \hline x&-3&-2&-1&0&1&2&3\\ \hline y&\frac{5}{2}&0&-\frac{3}{2}&-2&-\frac{3}{2}&0&\frac{5}{2}\\ \hline \end{array}\]

Now, we’ll plot those points on a coordinate plane:

Notice that this process is the same as graphing a function.

Constraints

There are times when every possible answer to an equation may not make sense in the real world. Other times, there may be given restrictions on possible answers. These are known as constraints, or limitations, on the quantities. These constraints can be represented with inequalities.

Suppose you have \(18\) marbles in a bag, and there are red and blue ones. You don’t know how many of each you have, but you know there are more red than blue marbles. Letting \(b\) equal the number of blue marbles and \(r\) the number of red marbles, we know two things:

\[b +r =18\] \[r>b\]

We can’t use the substitution or elimination method to figure out the values of the two variables because we don’t have two equalities. But we can figure out the constraints of the two variables. Let’s start by solving the equation for \(b\) and substitute that into the inequality:

\[b=18-r\] \[r>18-r\] \[2r>18\] \[r>9\]

We also know that \(r<18\) because there must be at least one blue marble, so the constraint is \(9<r<18\). This doesn’t tell us what the value of \(r\) is, but it does narrow it down. The number of red marbles is constrained to any number from \(10\) to \(17\).

We could find the constraints of blue marbles the same way.

Let’s try another example.

A rectangle’s length is three times its width. If its perimeter is less than \(24\), what are the constraints of its length and width?

Solution

We know two things. First, the length is three times as long the width, which we can represent as an equation:

\[l=3w\]

Second, since we are assuming the perimeter is less than \(24\), we can use the formula for perimeter and write:

\[2(l+w)<24\]

Now, let’s substitute \(3w\) in place of \(l\) in the inequality:

\[2(3w+w)<24\]

Solving for \(w\), we get:

\[8w<24\] \[w<3\]

That tells us that the width must be less than \(3\). Now, we’ll multiply both sides by \(3\) and substitute \(l\) for \(3w\):

\[3w<9\] \[l<9\]

We now know that \(l\) must be less than \(9\).

Those are the constraints on the length and width of the rectangle.

Linear Inequalities

The operations of equalities, like addition, subtraction, multiplication, and division, all work for inequalities too. The only difference is that if you multiply or divide both sides by a negative number, you have to change the direction of the inequality sign.

Let’s say you have this inequality:

\[5-2x <9\]

Just as if it were an equation, you’ll subtract the \(5\) from both sides and simplify:

\[5-2x-5<9-5\] \[-2x< 4\]

The last step is dividing both sides by -2, but because this is an inequality, you need to switch the direction of the sign from less than (\(<\)) to greater than (\(>\)):

\[x>-2\]

Solving in Terms of Another Variable

You may be asked to solve an equation in terms of another variable. For instance, you could be asked to solve \(5x-a=-3\) in terms of \(a\). The phrase “in terms of \(a\)” simply means \(a\) will appear in your answer. You won’t be able to get rid of the \(a\).

To do this, just follow the normal steps to get \(x\) alone on the left side. Whatever ends up on the right side is your answer:

\[5x-a=-3\] \[5x = a- 3\] \[x = \frac{a-3}{5}\]

Let’s do one more example problem, this time with more than one variable that is not \(x\).

Solve the following equation in terms of \(a\) and \(b\):

\[2ax+b=10+2b\]

Solution

Despite the extra variable, the process is exactly the same. We simply need to isolate \(x\), one variable at a time:

\[2ax=10 +2b -b\] \[2ax = 10 +b\] \[x = \frac{10+b}{2a}\]

That answer might not look very neat, but that’s as simplified as it can be.

Systems of Linear Equations

We solved systems of two linear equations earlier in this guide by adding or subtracting the equations to eliminate one of the variables, as well as by substituting values from one equation into the other. Another way to solve pairs of equations is by graphing both equations and seeing where they intersect. Their intersection point on a graph will have an \(x\) and a \(y\) value, just like we got with the elimination and substitution methods.

Consider this graph below. The equations for the two lines are \(y-0.6x=0\) and \(-y+x=3\). At what point do these lines intersect?

It’s a little bit hard to read, but the point of intersection is \((7.5, 4.5)\). That’s our answer. To verify that’s correct, let’s solve these equations for \(x\) and \(y\) using the elimination method. First, we’ll add the two equations so the \(y\) disappears:

\[\begin{array}{rcrcr} y &-&0.6x&=&0\\ -y &+&x&=&3\\ \hline & &0.4x&=&3\\ & &x& =&\frac{3}{0.4}\\ & &x& =&7.5\\ \end{array}\]

Knowing \(x=7.5\), we can substitute it into \(-y+x=3\):

\[-y+x=3\] \[-y+7.5=3\] \[-y=-4.5\] \[y=4.5\]

See, we got the same answer as we did when we graphed it.

Understanding Graphs

Suppose you have an equation in two variables, maybe something simple like \(y=2x-1\). How many \((x, y)\) pairs will make the equation true? Does \((1,1)\) work? Yes. How about \((2, 3)\)? It does. And \((-1, -3)\)? Yes, again. Try these for yourself to see that they all work.

We saw three pairs that work, but how many pairs exist that will work? If you are thinking it’s an infinite number, you are right. There are an infinite number of solutions to that equation, which corresponds to an infinite number of points on a graph of the equation. But how many of these points would we need to plot a graph of the equation? We certainly aren’t going to graph an infinite number of points. Luckily, we can see that this is a linear equation, so all we need is two points:

The line graph above consists of how many points? Are you thinking it’s an infinite number again? You’re correct. We only needed two points to locate it, but once it’s drawn it’s assumed to have an infinite number of points, going on beyond the visible portion of the graph. This is true of any graph.

The only other thing to note here is that if the equation made a curved graph, it would have taken more points to show its general shape, but you still wouldn’t have needed to graph every point. The graph (straight or curved) represents every point of its related equation.